solve this equation { ((xy+x+y=19)),((yz + y+z = 11)),((z+x+zx = 14)) :}


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 121973 by bemath last updated on 13/Nov/20

$s o l v e t h i s e q u a t i o n$ ${\begin{cases} x y + x + y = 19 \\ y z + y + z = 11 \\ z + x + z x = 14 \end{cases}$
Commented by liberty last updated on 13/Nov/20
$→ x(y+1)=19−y ∧ z(y+1)=11−y ⇒ { ((x=((19−y)/(y+1)))),((z=((11−y)/(y+1)))) :} ⇔ substitution to eq(3) give ((19−y)/(y+1)) + ((11−y)/(y+1)) + (((19−y)(11−y))/((y+1)^2 )) = 14 which is equivalent to 14(y+1)^2 −(y+1)(30−2y)−(19−y)(11−y)=0 simplifying to y^2 +2y−15=0 → { ((y_1 =3)),((y_2 =−5)) :} the corresponding values of x and z are computed from the previous formulas we get { ((x_1 =4 , x_2 =−6)),((z_1 =2 , z_2 =−4)) :} so there are exactly two solution triples (x,y,z): (4,3,2) and (−6,−5,−4).▲$
$\to x (y + 1) = 19 - y \land z (y + 1) = 11 - y$ $\Rightarrow {\begin{cases} x = \frac{19 - y}{y + 1} \\ z = \frac{11 - y}{y + 1} \end{cases} \Leftrightarrow substitution to eq (3)$ $give \frac{19 - y}{y + 1} + \frac{11 - y}{y + 1} + \frac{(19 - y) (11 - y)}{{(y + 1)}^{2}} = 14$ $which is equivalent to$ $14 {(y + 1)}^{2} - (y + 1) (30 - 2 y) - (19 - y) (11 - y) = 0$ $simplifying to y^{2} + 2 y - 15 = 0 \to {\begin{cases} y_{1} = 3 \\ y_{2} = - 5 \end{cases}$ $the corresponding values of x and z$ $are computed from the previous formulas$ $we get {\begin{cases} x_{1} = 4, x_{2} = - 6 \\ z_{1} = 2, z_{2} = - 4 \end{cases}$ $so there are exactly two solution$ $triples (x, y, z) : (4, 3, 2) and (- 6, - 5, - 4) . ▴$
Commented by MJS_new last updated on 13/Nov/20

$this is easy . it leads to a 2^{nd} degree polynome,$ $no problems at all$ $(x ∣ y ∣ z) = (- 6 ∣ - 5 ∣ - 4) \lor (4 ∣ 3 ∣ 2)$
Commented by bemath last updated on 13/Nov/20

$t h a n k y o u b o t h$
Commented by soumyasaha last updated on 13/Nov/20
$solve this equation { ((xy+x+y=19)),((yz + y+z = 11)),((z+x+zx = 14)) :} xy + x + y = 19 ⇒ xy + x + y +1= 19+1 ⇒ (1+x)(1+y)= 20.......(i) Simiilarly, (1+y)(1+z) = 12.......(ii) and (1+z)(1+x) = 15.......(iii) Multiplying (i), (ii), (iii) we get, (1+x)^2 (1+y)^2 (1+z)^2 = 20.12.15 Taking positive values, (1+x)(1+y)(1+z)= 60.........(iv) Dividing (iv) by (ii) (1+x) = 5 ⇒ x = 4 similarly, y = 3 and z = 2$
$s o l v e t h i s e q u a t i o n$ ${\begin{cases} x y + x + y = 19 \\ y z + y + z = 11 \\ z + x + z x = 14 \end{cases}$ $xy + x + y = 19$ $\Rightarrow xy + x + y + 1 = 19 + 1$ $\Rightarrow (1 + x) (1 + y) = 20 . . . . . . . (i)$ $Simiilarly, (1 + y) (1 + z) = 12 . . . . . . . (ii)$ $and (1 + z) (1 + x) = 15 . . . . . . . (iii)$ $Multiplying (i), (ii), (iii) we get,$ ${(1 + x)}^{2} {(1 + y)}^{2} {(1 + z)}^{2} = 20.12.15$ $Taking positive values,$ $(1 + x) (1 + y) (1 + z) = 60 . . . . . . . . . (iv)$ $Dividing (iv) by (ii)$ $(1 + x) = 5 \Rightarrow x = 4$ $similarly, y = 3 and z = 2$
	Answered by nimnim last updated on 13/Nov/20
	${ ((xy+x+y=19)),((yz+y+z=11)),((z+x+zx=14)) :}⇒ { (((x+1)(y+1)=20.....(1))),(((y+1)(z+1)=12.....(2))),(((z+1)(x+1)=15.....(3))) :} ⇒(x+1)^2 (y+1)^2 (z+1)^2 =20.12.15 ⇒(x+1)(y+1)(z+1)=±60.......(4) solving (1),(2),(3) and (4) we get (x+1)=±5, (y+1)=±4, (z+1)=±3 ⇒x=4,−6, y=3,−5, z=2,−4 ∴(x,y,z)= (4,3,2)∨ (−6,−5,−4)$
	${\begin{cases} x y + x + y = 19 \\ y z + y + z = 11 \\ z + x + z x = 14 \end{cases} \Rightarrow {\begin{cases} (x + 1) (y + 1) = 20 . . . . . (1) \\ (y + 1) (z + 1) = 12 . . . . . (2) \\ (z + 1) (x + 1) = 15 . . . . . (3) \end{cases}$ $\Rightarrow {(x + 1)}^{2} {(y + 1)}^{2} {(z + 1)}^{2} = 20.12.15$ $\Rightarrow (x + 1) (y + 1) (z + 1) = \pm 60 . . . . . . . (4)$ $s o l v i n g (1), (2), (3) a n d (4) w e g e t$ $(x + 1) = \pm 5, (y + 1) = \pm 4, (z + 1) = \pm 3$ $\Rightarrow x = 4, - 6, y = 3, - 5, z = 2, - 4$ $∴ (x, y, z) = (4, 3, 2) \lor (- 6, - 5, - 4)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum